How do you integrate $(ln (x) \cdot (\frac{1}{x})) d x$ $(ln(x)*(1/x))dx$ ?

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Answer 1 · 2016-07-03T17:03:11

$= \frac{1}{2} {(ln x)}^{2} + C$ $= 1/2 (ln x)^2 + C$

$int dx qquad ln(x)*1/x$

you can do this by inspection as $(ln x) ' = 1/x$ so we can trial $alpha (ln x)^2$ as a solution.

So $(alpha (ln x)^2 + C)' = 2 alpha ln x 1/x implies 2 alpha = 1, alpha = 1/2$

if you don't fancy that you could use IBP : $int uv' = uv - int u'v$

$u = ln x, u' = 1/x$
$v' = 1/x, v = ln x$

$implies color{red}{int dx qquad ln(x)1/x} = (ln x)^2 - color{red}{int dx qquad ln x 1/x }+ C$

$implies 2 int dx qquad ln(x)1/x = (ln x)^2 + C$

$implies int dx qquad ln(x)1/x = 1/2 (ln x)^2 + C$

How do you integrate (ln(x)*(1/x))dx(ln(x)⋅(1x))dx(ln(x)*(1/x))dx?